Question

用节点电压法求如图各支路的电流

Answer 1

节点电压方程：(1/5+1/6+1/2)Uab=10/5-12/2
解得Uab=-4.6154V
I1=(10+4.6154)/5=2.923A
I2=-4.6154/6=-0.769A
I3=(12-4.6154)/2=3.692A

Answer 2

(1/5+1/6+1/2 )Uab=10/5-12/2=-4
Uab=-4/(1/5+1/6+1/2 )V
I1=(10-Uab)/5
I2=(Uab+12)/2
I3=-Uab/6